My problem is the following :\
Show that $4^m + 11^n = x^2$ with $(m,n,x) \in \mathbb{N}^3$ has no solution.
My first idea is to separate the problem into two cases:
when $n$ is even, and when $n$ is odd (I already used this method in similar problems).
Then, I guess you have to look at the modulo 4 problem, but it doesn't give anything.
Does anyone have any idea how to solve this problem ?
Lucas.
We have $x^2-4^m=(x-2^m)(x+2^m)=11^n$. Both $(x-2^m)$ and $(x+2^m)$ cannot be divisible by $11$ because their difference is a power of $2$. So the smaller of them is $1$, and the bigger is $11^n$: $x-2^m=1, x+2^m=11^n$. So $2^{m+1}+1=11^n$. So $11^n-1=2^{m+1}$. Hence $2^{m+1}$ is divisible by $5$, a contradiction.